Asked by Dante
a rancher has 1200 fet of fencing to enclose two adjacent cattle pens. What dimensions should be used so that the enclosed area is maximized?
Answers
Answered by
Reiny
You don't describe the diagram, but I am familiar with this type of question.
Assuming there is a common width between the two fields
let the width be x ft
let the length of the combined field be y
so
3x + 2y = 1200
y = 600 - (3/2)x
area = xy
= x(600 - (3/2)x)
= 600x - (3/2)x^2
d(area)/dx = 600 - 3x
= 0 for a max of area
3x = 600
x = 200
y = 600-(3/2)(200) = 300
state the conclusion
Assuming there is a common width between the two fields
let the width be x ft
let the length of the combined field be y
so
3x + 2y = 1200
y = 600 - (3/2)x
area = xy
= x(600 - (3/2)x)
= 600x - (3/2)x^2
d(area)/dx = 600 - 3x
= 0 for a max of area
3x = 600
x = 200
y = 600-(3/2)(200) = 300
state the conclusion
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